Question

A block of mass m is attached to two unstretched springs of spring constant k₁ and k₂ as shown in figure . The block is displaced towards right through a distance of x and released. Find the speed of the block as it passes through the mean position shown.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Assume that the speed of the block = v 
Displacement in the body = x 

Now for the potential energy , which is stored in the spring at the time of compression .
Potential Energy , P.E = 1/2k₁x₁² + 1/2k₂x₂²

In the mean position, by law of conservation of energy 
So, Potential energy changes to Kinetic energy of the block.
1/2mv² = 1/2k₁x₁² +1/2k₂x₂² 
mv² = x² (k₁ + k₂)
v² = x² (k₁ +k₂ ) /m
$v = ( \sqrt \frac{{k_{1} +k_{2}} }{m} )$ .x  


Hence the speed of the block when it passes through their mean position is 
$v = ( \sqrt \frac{{k_{1} +k_{2}} }{m} )$ .x  

Hope it helps. 

Answer 2

see the attachment for the answer